I'm trying to figure out a point(D) on B-C that is a specific distance from A.
I really don't have any idea on how to proceed with the ellipsoidal math so I figured I'd ask here. How would I figure out the longitude, latitude of D which is 5mi(26400ft) away from A, on the line B-C? The bearing of A-D would be just as useful as i could figure out the point 5mi away from A with the bearing.

I wouldn't mind if the formula lined up with the Vincenty model of the earth.

2 Answers
2

This is most easily solve using the equidistant azimuthal projection.
Guess some point for D, e.g., A. Transform A, B, C, to equidistant
azimuthal projection using D as a center. In projected space, solve for
D' (i.e., D' lies on BC and is a distance s from A). Project D' back to
lat,lon and update D with this position. Repeat. This converges
quadratically.

This ends up being really close, the distance is correct but It doesn't land on line segment B-C?
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PhilMay 26 '13 at 17:41

Yes! The distance is right because the projection is equidistant (with D as the center). D lies on B-C because the projection is azimuthal (with D as the center). I should add that you can pretty much choose any point as a starting point: A, B, C, (B+C)/2, the North Pole, the Eiffel Tower all work fine.
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cffkMay 28 '13 at 12:42

the distance is right.. but how would you go about getting a point between B & C. instead of just on the line B-C.
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PhilMay 31 '13 at 4:05

My apologies, I got you the solution outside the range B-C. Change the sign of the sqrt in planesolve.m and rerun and you will get the solution you want 41.1687907834032 -86.2525504601801. In general you should modify the code to try both solutions in planesolve and then select the one (if any) which gives you opposite azimuths D->B and D->C.
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cffkMay 31 '13 at 9:36